Question

4log0,6 x>= log0,6 8+log0,6 2

Answer 1

Log0,6x в 4 степени больше=log0,6 16,т.к     0 больше 0,6 но меньше 1, то
х в 4 степени меньше = 16 тогда -2 больше=х меньше = 2

Answer 2

To solve the inequality 4 * log₀.₆(x) ≥ log₀.₆(8) + log₀.₆(2), we can use the properties of logarithms and some algebraic manipulation.

First, let's simplify the right side of the inequality by combining the two logarithms using the properties of logarithms:

4 * log₀.₆(x) ≥ log₀.₆(8 * 2)

4 * log₀.₆(x) ≥ log₀.₆(16)

Now, we can express the right side logarithm in terms of the same base (0.6):

4 * log₀.₆(x) ≥ log₀.₆(0.6²⁵)

4 * log₀.₆(x) ≥ 5

Next, we want to isolate the logarithmic term. Divide both sides of the inequality by 4:

log₀.₆(x) ≥ 5/4

Now, to remove the logarithm, we can rewrite this inequality in exponential form. Recall that log base a of b is equivalent to a raised to the power of b:

x ≥ 0.6^(5/4)

Now, we can calculate the right side of the inequality:

x ≥ 0.6^(5/4) ≈ 0.776

So, the solution to the inequality is x ≥ 0.776.